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Consider the vector field $\mathbf F = \langle x+x^3e^y, -3x^2e^y \rangle$ and let $C$ be the circle $x^2 +y^2=5^2$. Let $\mathbf u$ be the unit vector $\dfrac{1}{\sqrt{x^2+y^2}}\langle x,y\rangle$ and $f(x,y) = \mathbf F \cdot \mathbf u$. Evaluate the line integral: $$\int_C f(x,y) \,\mathrm{d}s,$$ where $\mathrm{d}s=\Vert\mathbf r'(t)\Vert \,\mathrm{d}t$.

So I noticed that this is a integral for a scalar field, which then I decided to parameterise the curve $C:\mathbf r(t)=\langle 5\cos(t),5\sin(t)\rangle,t\in[0,2\pi]$, then $\mathbf r'(t)=\langle -5\sin(t),5\cos(t)\rangle $ and $\Vert\mathbf r'(t)\Vert=5$. So,$$ \int_C f(x,y)\,\mathrm{d}s = \int_{0}^{2\pi}\frac{1}{\sqrt{x^2+y^2}}\langle x,y\rangle\cdot\langle x+x^3e^y, -3x^2e^y\rangle \Vert\mathbf r'(t)\Vert\,\mathrm{d}t.$$

Simplification yields:$$ \int_{0}^{2\pi}(25\cos^2(t)+625\cos^4(t)e^{5\sin(t)} -375\cos^2(t)\sin(t)e^{5\sin(t)}) \,\mathrm{d}t,$$ which is kind of impossible to integrate.

I am not sure if there is an easier way to do this question. Upon checking the answer I noticed that Green's theorem was used, I am not sure on how Green's theorem is applicable to a line integral on a scalar field. (I was only thought on how to apply it onto a vector field). Any help on this clarification is really appreciated!

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$\def\vec{\boldsymbol}\def\d{\mathrm{d}}\def\e{\mathrm{e}}$Suppose $D = \{(x, y) \in \mathbb{R}^2 \mid x^2 + y^2 \leqslant 5\}$, then $\partial D = C$. Because $\vec{u}$ is the outer-pointing normal along path $C$, then$$ \vec{u} \,\d s = (\d y, -\d x), $$ and for $\vec{F} = (P, Q) = (x + x^3 \e^y, -3x^2 \e^y) $,$$ \int\limits_C f(x, y) \,\d s = \int\limits_C \vec{F} · \vec{u} \,\d s = \int\limits_C (-Q \,\d x + P \,\d y). $$ By Green's theorem,\begin{align*} &\mathrel{\phantom{=}}{} \int\limits_C (-Q \,\d x + P \,\d y) = \iint\limits_D \left( \frac{\partial P}{\partial x} - \left( -\frac{\partial Q}{\partial y} \right) \right) \,\d x \d y\\ &= \iint\limits_D ((1 + 3x^2 \e^y) - 3x^2 \e^y) \,\d x \d y = \iint\limits_D \d x \d y = 25π. \end{align*}

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  • $\begingroup$ Thanks for the help! Could I ask how did you obtain u ds = (dy, -dx)? I am not sure how this works. $\endgroup$ – Derp Apr 25 '18 at 14:38
  • $\begingroup$ @Derp In this specific case, this step could be rigorously justified by using the parameterization of $C$ and the parameterized representation of line integral. In fact, as long as $C$ is a $C^1$ curve, this step holds. $\endgroup$ – Saad Apr 25 '18 at 14:55

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